Permutations and CombinationsMCQPYQ Dec. 21Question 1615 of 251
All Questions

The number of four-letter words can be formed using the letters of the word DICTIONARY is

Options

A5040
B720
C90
D30240
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option a — 5040

All Options:

  • A5040
  • B720
  • C90
  • D30240

Ad

Detailed Solution & Explanation

In the word **DICTIONARY**, there are 10 letters: D, I, C, T, I, O, N, A, R, Y. Although 'I' is repeated twice, the standard textbook solution treats all 10 letters as distinct entities.

Under this assumption:
- Total letters = 10.
- Number of letters in the word to be formed = 4.
- Total number of 4-letter words = 10P4\displaystyle ^{10}P_4:
10P4=10times9times8times7=5040^{10}P_4 = 10 \\times 9 \\times 8 \\times 7 = 5040
Let us also perform the mathematically rigorous derivation that accounts for the repeated letter 'I':
The distinct letters are {D,I,C,T,O,N,A,R,Y}\displaystyle \{D, I, C, T, O, N, A, R, Y\} (9 distinct, with 'I' having frequency 2, and others frequency 1).
- **Case 1: All 4 letters are distinct.** We select 4 distinct letters from 9: 9P4=3024\displaystyle ^9P_4 = 3024.
- **Case 2: 2 letters are 'I' and the other 2 are distinct.** We select 2 other distinct letters from 8: 8C2timesfrac4!2!=28times12=336\displaystyle ^8C_2 \\times \\frac{4!}{2!} = 28 \\times 12 = 336.
- **Total rigorous ways:** 3024+336=3360\displaystyle 3024 + 336 = 3360.
Since 3360 is not in the options, the textbook assumes all letters are distinct.
Hence, **Option A** is the correct answer.

About This Chapter: Permutations and Combinations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Factorials, Permutations, Combinations

This chapter deals with the fundamental principles of counting. It covers factorials, circular permutations, restricted permutations, combinations, and the differences between selecting items versus arranging them.

View Official ICAI Syllabus

Exam Strategy Tip

The most common mistake is confusing 'P' (Arrangement) with 'C' (Selection). If order matters (like opening a lock), use P. If order doesn't matter (like choosing a team), use C.

Related Comparison Tables

Permutations vs Combinations: The Quantitative Aptitude Guide

More Questions from Permutations and Combinations

The value of N\displaystyle N in N!+17!=18!\displaystyle N! + \frac{1}{7!} = \frac{1}{8!} is

A person can go from place 'A' to 'B' by 11 different modes of transport but is allowed to return to 'A' by any mode other than the one earlier. The number of different ways in which the entire journey can be completed is:

If a man travels from place A to B in 10 ways then by how many ways can he come back by another train?

If n!10(n−1)!=(n−1)!(n−3)!\displaystyle \frac{n!}{10(n-1)!} = \frac{(n-1)!}{(n-3)!} find 'n'.

Which of the following is a correct statement.

nP3=2⋅nP2\displaystyle ^nP_3 = 2 \cdot ^nP_2. Find n\displaystyle n.

Ready to Master Permutations and Combinations?

Practice all 251 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free